What is $b_n$ (a priori we might have $b_n=0$ for all $n$)? Do you really talk about a set of numbers $e^{b_n z}$ for fixed $z$, or do you rather mean the set of fucntions $z\mapsto e^{b_n z}$?
–
Hagen von EitzenSep 5 '12 at 15:09

@Zeraoulia I edited your post to make it a little clearer. Please look it over to make sure I didn't make any mistakes.
–
axblountSep 5 '12 at 15:21

1

It would still be helpful to know about which vector space over which field we are talking (and what we know about $b_n$)
–
Hagen von EitzenSep 5 '12 at 15:31

1 Answer
1

Pending clarification from OP of the matters Hagen raised in the comments, these remarks may be useful.

Suppose $b_n=n$ for all $n$, and take $z=1$, so in the first question we are talking about the numbers $e,e^2,e^3,\dots$. That these numbers are linearly independent over the rationals is another way of saying that $e$ is a transcendental number. The transcendence of $e$ is well-known and not at all trivial to prove.

Given one complex number $z=z_0$ such that the numbers $e^{b_nz}$ are linearly independent over the field of your choice, it is immediate that the functions $e^{b_nz}$ are linearly independent over that field, since any linear combination of the functions will fail to vanish when evaluated at $z=z_0$.

The set $b_n=b_n(z)$ is sequences of functions in the variable $z$. I talk about a set of numbers $\exp(b_n(z))$ for fixed $z$ in the first question and for the set of fucntions $z\mapsto\exp(b_n(z))$ in the second one. Here $z$ is a complex number and the field is also the complex plane.
–
user39552Sep 7 '12 at 14:02